15pts求调，悬赏2关+10rmb

# include <bits/stdc++.h>
using namespace std;
const int N = 301;
const int M = 2e6 + 7;
int T, n, m, k;
int a[M];
int t[N][2];
int op[M << 1]; 
int ans[M << 1][2];
int cnt;
int to[N << 1]; int tot;
int cntk; int p[N << 1];
inline void work( int pos, int i ) { // 处理 k = 2n - 2 的情况
	if(t[pos][1] == a[i]) op[++cnt] = 1, ans[cnt][0] = pos, t[pos][1] = 0;
	else if(t[pos][1] > 0) {
			op[++cnt] = 1, ans[cnt][0] = n;
			op[++cnt] = 2, ans[cnt][0] = pos, ans[cnt][1] = n;
			t[pos][0] = t[pos][1], t[pos][1] = 0;
		}
	else if(t[pos][0] == a[i]) op[++cnt] = 1, ans[cnt][0] = pos, t[pos][0] = 0, cntk ++;
	else if(t[pos][0] > 0) op[++cnt] = 1, ans[cnt][0] = pos, t[pos][1] = a[i]; 
	else op[++cnt] = 1, ans[cnt][0] = pos, t[pos][0] = a[i], cntk --;
}

int main() {
	ios :: sync_with_stdio(false); cin.tie(0), cout.tie(0);
	cin >> T;
	while ( T -- ) {
	//	memset(t, 0, sizeof t);
		cnt = 0; tot = 0;
		cin >> n >> m >> k;
		for ( int i = 1; i <= n; i ++) t[i][0] = t[i][1] = 0;
		for ( int i = 1; i <= n; i ++) to[i] = 0;
		for ( int i = 1; i <= m; i ++ ) cin >> a[i];
		if ( k == n * 2 - 2 ) {
			for ( int i = 1; i <= m; i ++ ) {
				int pos = a[i] + 1 >> 1;
				work(pos, i);
			}
		}
		if ( k == n * 2 - 1 ) {
			cntk = n;
			for ( int i = 1; i <= m; i ++ ) if ( to[a[i]] == 0 ) to[a[i]] = (++ tot + 1) >> 1;
			for ( int i = 1; i <= m; i ++ ) {
				if ( cntk == n ) { // 我的思路是，每当序列为空的时候，找到第一个需要 k = 2n - 1 策略的点，重新分配编号。这里不太会实现呀....
					memset(p, 0, sizeof p); int cntp = 0; int kk = 0, kkk = 0;
					for ( int j = i; j <= m; j ++ ) {
						p[a[j]] ^= 1;
						if (to[a[j]] == n) kkk = a[j];
						if (cntp == k) {
							kk = a[j];
							break;
						}
						cntp = p[a[j]] > 0 ? cntp + 1 : cntp - 1;
					}
					if(kk && kkk) swap(to[kk], to[kkk]);
					cout << "kk : " << kk << " " << kkk << "\n";
				}
				if ( to[a[i]] != n ) work(to[a[i]], i);
				if ( to[a[i]] == n ) {
					if ( t[n][0] > 0 ) {
						op[++cnt] = 1, ans[cnt][0] = n; t[n][0] = 0; cntk ++;
						continue;
					}
					if ( a[i + 1] == a[i] ) {
						op[++cnt] = 1, ans[cnt][0] = n;
						op[++cnt] = 1, ans[cnt][0] = n;
						i ++;
						continue;
					}
					int pos = to[a[i + 1]];
					if ( t[pos][1] == a[i + 1] ) {
						op[++cnt] = 1, ans[cnt][0] = n; t[n][0] = a[i]; cntk --;
						op[++cnt] = 1, ans[cnt][0] = pos;
						t[pos][1] = 0;
						i ++;
						continue;
					}
					if ( t[pos][0] == a[i + 1] && t[pos][1] == 0 ) {
						op[++cnt] = 1, ans[cnt][0] = n; t[n][0] = a[i]; cntk --;
						op[++cnt] = 1, ans[cnt][0] = pos;
						t[pos][0] = 0; cntk ++;
						i ++;
						continue;
					}
					if ( t[pos][0] == 0 && t[pos][1] == 0 ) {
						op[++cnt] = 1, ans[cnt][0] = n; t[n][0] = a[i]; cntk --;
						op[++cnt] = 1, ans[cnt][0] = pos;
						t[pos][0] = a[i + 1]; cntk --;
						i ++;
						continue;
					}
					if ( t[pos][0] > 0 && t[pos][0] != a[i + 1] ) {
						op[++cnt] = 1, ans[cnt][0] = n; t[n][0] = a[i]; cntk --;
						op[++cnt] = 1, ans[cnt][0] = pos;
						t[pos][1] = a[i + 1];
						i ++;
						continue;
					}
					if ( t[pos][0] == a[i + 1] && t[pos][1] > 0 ) {
						op[++cnt] = 1, ans[cnt][0] = pos;
						op[++cnt] = 1, ans[cnt][0] = n;
						op[++cnt] = 2, ans[cnt][0] = pos, ans[cnt][1] = n;
						t[pos][0] = t[pos][1], t[pos][1] = a[i]; cntk --;
						swap(to[a[i]], to[a[i + 1]]);
						i ++;
					}
				}
			}
		}
		cout << cnt << "\n";
		for ( int i = 1; i <= cnt; i ++ ) {
			if ( op[i] == 1 ) cout << op[i] << " " << ans[i][0] << "\n";
			if ( op[i] == 2 ) cout << op[i] << " " << ans[i][0] << " " << ans[i][1] << "\n";
		}
	}
	return 0;
}